# Posted by COFFEE on Monday, July 9, 2007 at 9:10pm.

Solve the initial-value problem.
y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4
r^2+4r+6=0,
r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1)
r=(16 +/- Sqrt(-8))
r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2)
y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2
y'(0)=4, c2=4
y(x)=e^(8x)*(2*cos(Sqrt(2)x)+4*sin(Sqrt(2)x))

What am I doing wrong here? This is the answer that I came up with but it is incorrect. Thanks.

* Calculus - Second Order Differential Equations - bobpursley, Monday, July 9, 2007 at 10:09pm

your use of the quadratic formula is wrong. r= (-b +- sqrt (b^2 -4ac)/2a

you did not use -b.
--------------------------
y''+4y'+6y=0, y(0)=2, y'(0)=4
r^2+4r+6=0, r=(-4 +/- sqrt(16-4(1)(6))/2
r=-2 +/- sqrt(2)*i
y=e^-2x*(c1*cos(sqrt(2))x+c2*sin(sqrt(2))x)
y(0)=1*(c1+0)=2, c1=2
y'=(-1/2)e^-2x*(c1*(sin(sqrt(2)))/sqrt(2)-c2*(cos(sqrt(2)))/sqrt(2))
y'(0)=(-1/2)(0-1/sqrt(2)*c2)=4
c2=2/sqrt(2)
y(x)=e^-2x*(2cos(sqrt(2))x+(2/sqrt(2))sin(sqrt(x))x)

What is wrong with my solution? thanks.

## In your initial calculations, when you used the quadratic formula, you made a mistake. The correct quadratic formula is:

r = (-b ± sqrt(b^2 - 4ac)) / 2a

But in your solution, you missed the negative sign in front of the coefficient of the linear term. You wrote it as r=(16 ± Sqrt(-8)), but it should be r=(-4 ± Sqrt(16-4(1)(6))/2. This error affected the subsequent steps of your solution and led to the incorrect answer.

Let's go through the correct solution step by step:

1. Start with the given differential equation: y'' + 4y' + 6y = 0.

2. Find the roots of the characteristic equation: r^2 + 4r + 6 = 0.

Use the quadratic formula to solve for r:
r = (-b ± sqrt(b^2 - 4ac)) / 2a.
In this case, a = 1, b = 4, and c = 6.
Plugging these values into the quadratic formula, we get:
r = (-4 ± sqrt(16 - 4(1)(6))) / 2
r = (-4 ± sqrt(16 - 24)) / 2
r = (-4 ± sqrt(-8)) / 2
r = (-4 ± sqrt(8)i) / 2
r = -2 ± sqrt(2)i.

3. Write the general solution for y(x):
Using the complex roots, r = -2 ± sqrt(2)i, we can write the general solution as:
y(x) = e^(-2x)(c1*cos(sqrt(2)x) + c2*sin(sqrt(2)x)).

4. Apply the initial conditions to find the particular solution:
y(0) = 2:
Plugging x = 0 and y = 2 into the general solution, we get:
2 = e^(-2*0)(c1*cos(0) + c2*sin(0))
2 = e^0(c1 + 0)
2 = c1
Therefore, c1 = 2.

y'(0) = 4:
Plugging x = 0 and y' = 4 into the derivative of the general solution, we get:
4 = (-2)e^(-2*0)(2*(sin(sqrt(2)*0))/sqrt(2) - c2*(cos(sqrt(2)*0))/sqrt(2))
4 = -2(0 - c2/sqrt(2))
4 = 0 + c2/sqrt(2)
c2 = 4*sqrt(2).

5. Substitute the values of c1 and c2 back into the general solution:
y(x) = e^(-2x)(2*cos(sqrt(2)x) + (4*sqrt(2))*sin(sqrt(2)x)).

So, the correct solution to the initial-value problem is:
y(x) = e^(-2x)(2*cos(sqrt(2)x) + (4*sqrt(2))*sin(sqrt(2)x)).

I hope this clears up the confusion and helps you understand the correct solution. Remember to always double-check your calculations to avoid mistakes.